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Data publicarii: 04 Noiembrie, 2014

EXERCITIUL 8

Suport teoretic:

Arii,functiile arcsin,arccos,integrale definite.

Enunt:

Sa se afle aria A a suprafetei delimitata de curbele reprezentative ale functiilor

f,g:[-1,+1] - > R, unde

f(x) = arcsinx

si

g(x) = arccosx

si dreptele de ecuatii

x=0,\;x=\frac{\sqrt{3}}{2}.x=0,\;x=\frac{\sqrt{3}}{2}.

Raspuns:

A=\frac{\pi\sqrt{3}-24\sqrt{2}+36}{12}.A=\frac{\pi\sqrt{3}-24\sqrt{2}+36}{12}.

Rezolvare:

A=\int_{0}^{\frac{\sqrt{3}}{2}}{|{arcsinx}-{arccosx}|}{dx}=\int_{0}^{\frac{\sqrt{3}}{2}}{|\frac{\pi}{2}-2{arcsinx}|}{dx},A=\int_{0}^{\frac{\sqrt{3}}{2}}{|{arcsinx}-{arccosx}|}{dx}=\int_{0}^{\frac{\sqrt{3}}{2}}{|\frac{\pi}{2}-2{arcsinx}|}{dx},

in baza identitatii cunoscute:

{arcsinx+arccosx}=\frac{\pi}{2},\;\forall{x}\in{[-1,1]}.{arcsinx+arccosx}=\frac{\pi}{2},\;\forall{x}\in{[-1,1]}.

Tinand cont de monotonia functiei arcsinx si de definitia modulului, avem:

\int_{0}^{\frac{\sqrt{3}}{2}}{|\frac{\pi}{2}-2{arcssinx}|}{dx}=\int_{0}^{\frac{\sqrt{3}}{2}}{|\frac{\pi}{2}-2{arcssinx}|}{dx}= \int_{0}^{\frac{\sqrt{2}}{2}}{(\frac{\pi}{2}-2{arcsinx})}{dx}+\int_{0}^{\frac{\sqrt{2}}{2}}{(\frac{\pi}{2}-2{arcsinx})}{dx}+ \int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}{(2{arcsinx}-\frac{\pi}{2})}{dx}.\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}{(2{arcsinx}-\frac{\pi}{2})}{dx}.

Deci: 

A={\frac{\pi}{2}}\cdot{\int_{0}^{\frac{\sqrt{2}}{2}}{dx}}-2\int_{0}^{\frac{\sqrt{2}}{2}}{arcsinx}{dx}+2\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}{arcsinx}{dx}-A={\frac{\pi}{2}}\cdot{\int_{0}^{\frac{\sqrt{2}}{2}}{dx}}-2\int_{0}^{\frac{\sqrt{2}}{2}}{arcsinx}{dx}+2\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}{arcsinx}{dx}- {\frac{\pi}{2}}\cdot{\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}{dx}}.{\frac{\pi}{2}}\cdot{\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}{dx}}.

Cum 

\int{arcsinx}{dx}=\int{x^{\int{arcsinx}{dx}=\int{x^{'}}{arcsinx}{dx}={xarcsinx}+\int{\frac{{(1-x^2)}^{'}}{2\sqrt{1-x^2}}}{dx}=

{xarcsinx}+\sqrt{1-x^2}+\mathcal{C},{xarcsinx}+\sqrt{1-x^2}+\mathcal{C},

rezulta:

A={\frac{\pi}{2}}x|_{0}^{\frac{\sqrt{2}}{2}}-2({xarcsinx}+\sqrt{1-x^2})|_{0}^{\frac{\sqrt{2}}{2}}+2({xarcsinx}+\sqrt{1-x^2})|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}-\frac{\pi}{2}x|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}.A={\frac{\pi}{2}}x|_{0}^{\frac{\sqrt{2}}{2}}-2({xarcsinx}+\sqrt{1-x^2})|_{0}^{\frac{\sqrt{2}}{2}}+2({xarcsinx}+\sqrt{1-x^2})|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}-\frac{\pi}{2}x|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{3}}{2}}.

In urma catorva calcule de rutina, se obtine in final:

A=\frac{\pi\sqrt{3}-24\sqrt{2}+36}{12}.A=\frac{\pi\sqrt{3}-24\sqrt{2}+36}{12}.


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